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Oksana_A [137]
3 years ago
14

I need help with this question answer ASAP

Mathematics
1 answer:
Sedaia [141]3 years ago
3 0

Answer:

A

Step-by-step explanation:

Plug 2 in for y in the second equation. You get 3 as x. So x=3 and y=2. (3,2)

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Use the first three triominoes to find the missing number in the fourth triomino.
Nonamiya [84]
Answer: 28

Since we know there is a pattern of 20, 22, 24, and 26, we can see that there is a difference between each of the numbers of +2. Therefore, 26 + 2 = 28

- Hope it helps!
4 0
3 years ago
What is the apparent solution to the system of equations?
Alexus [3.1K]
X = 3
y = -3

You will see this using any graphing tool at your disposal. 
7 0
4 years ago
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rusak2 [61]

Answer:

A

Step-by-step explanation:

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4 0
2 years ago
Read 2 more answers
In a string of 12 Christmas tree light bulbs, 3 are defective. The bulbs are selected at random and tested, one at a time, until
Juliette [100K]

Using the hypergeometric distribution, it is found that there is a 0.0273 = 2.73% probability that the third defective bulb is the fifth bulb tested.

In this problem, the bulbs are chosen without replacement, hence the <em>hypergeometric distribution</em> is used to solve this question.

<h3>What is the hypergeometric distribution formula?</h3>

The formula is:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • N is the size of the population.
  • n is the size of the sample.
  • k is the total number of desired outcomes.

In this problem:

  • There are 12 bulbs, hence N = 12.
  • 3 are defective, hence k = 3.

The third defective bulb is the fifth bulb if:

  • Two of the first 4 bulbs are defective, which is P(X = 2) when n = 4.
  • The fifth is defective, with probability of 1/8, as of the eight remaining bulbs, one will be defective.

Hence:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}

P(X = 2) = h(2,12,4,3) = \frac{C_{3,2}C_{9,1}}{C_{12,4}} = 0.2182

0.2182 x 1/8 = 0.0273.

0.0273 = 2.73% probability that the third defective bulb is the fifth bulb tested.

More can be learned about the hypergeometric distribution at brainly.com/question/24826394

8 0
2 years ago
12.3 Write the first five terms of the sequence whose general term is <img src="https://tex.z-dn.net/?f=a_%7Bn%7D%3D3%5E%7Bn%7D%
weeeeeb [17]

Answer:

7,13,31,85,247......

Step-by-step explanation:

a_n =3^n + 4

        \sf a_1=3^1 + 4 = 3 + 4 = 7\\\\a_2 = 3^2 + 4 = 9 + 4 = 13\\\\a_3 = 3^3 + 4 = 27 + 4 = 31\\\\a_4 = 3^4  + 4 = 81 + 4 = 85\\\\a_5 = 3^5 + 4 = 243 + 4 = 247

8 0
2 years ago
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